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describe gravitation using a field model and apply Newton's law of universal gravitation F=Gm1m2/r2F = G m_1 m_2 / r^2 and the relationships g=GM/r2g = G M / r^2, g=F/mg = F/m, the work done by a gravitational field W=ΔU=mgΔhW = \Delta U = mg \Delta h in a uniform field and the change in gravitational potential energy in non-uniform fields as the area under a force-distance graph

A focused answer to the VCE Physics Unit 3 dot point on gravitational fields. Covers the field model and field lines, Newton's law of universal gravitation, the equivalence of gg as field strength and as acceleration, gravitational potential energy in uniform and non-uniform fields, and how to read change in UU as the area under a FF vs rr graph.

Generated by Claude Opus 4.812 min answer

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  1. What this dot point is asking
  2. The answer
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What this dot point is asking

VCAA wants you to describe gravity using a field model, apply Newton's law of universal gravitation, calculate field strength gg, and find changes in gravitational potential energy both in a uniform field (mgΔhmg\Delta h) and in a non-uniform field (the area under the FF vs rr graph).

The answer

The field model

A gravitational field surrounds every mass. A second mass placed in the field experiences a force toward the first. The field is a vector at each point, drawn with field lines that point in the direction of the force on a positive test mass.

  • Near a planet's surface the field is approximately uniform (parallel field lines, g9.8g \approx 9.8 N/kg downward).
  • At large distances the field is radial (lines pointing inward toward the planet) and the strength falls as 1/r21/r^2.

Newton's law of universal gravitation

Every pair of masses attracts with a force:

F=Gm1m2r2F = \frac{G m_1 m_2}{r^2}

where G=6.67×1011G = 6.67 \times 10^{-11} N m squared per kg squared and rr is the centre-to-centre distance.

Field strength

The gravitational field strength gg at a point is the force per unit mass:

g=Fm=GMr2g = \frac{F}{m} = \frac{G M}{r^2}

It is also numerically equal to the acceleration of a freely falling object (because F=mgF = mg and F=maF = ma together give a=ga = g). Units are N/kg or m/s squared.

At Earth's surface, g=9.8g = 9.8 N/kg. The field strength falls with altitude as 1/r21/r^2 where rr is measured from Earth's centre.

Gravitational potential energy in a uniform field

Close to a planet's surface, the field is approximately uniform. The change in gravitational potential energy when an object of mass mm is raised through height Δh\Delta h is:

ΔU=mgΔh\Delta U = m g \Delta h

This is the work done against gravity.

Gravitational potential energy in a non-uniform field

Far from a planet's surface, the field varies with distance and the equation mgΔhmg\Delta h no longer applies. The change in potential energy between two distances r1r_1 and r2r_2 equals the work done against the gravitational force, which is the area under the FF vs rr graph between those points.

VCAA expects you to use the graphical area for non-uniform fields. The analytic equivalent is:

ΔU=GMm(1r21r1)=GMm(1r11r2)\Delta U = -G M m \left( \frac{1}{r_2} - \frac{1}{r_1} \right) = G M m \left( \frac{1}{r_1} - \frac{1}{r_2} \right)

for moving from r1r_1 to r2>r1r_2 > r_1 (positive when moving outward, because work is done against gravity).

Field lines and energy: a quick visual rule

  • Field lines point in the direction of the force on a positive test mass (always toward the planet).
  • A mass moved with the field (downward) loses potential energy.
  • A mass moved against the field (outward) gains potential energy.

Examples in context

Example 1. ANU Mt Stromlo satellite orbit determination. Mt Stromlo's satellite-laser-ranging station tracks the 99 tonne LAGEOS satellite at altitude 59005900 km, orbital radius r=6378+5900=12280r = 6378 + 5900 = 12\,280 km. Newton's law gives gravitational field strength at this radius: g=GM/r2=6.674×1011×5.97×1024/(1.228×107)2=2.64g = GM/r^2 = 6.674 \times 10^{-11} \times 5.97 \times 10^{24}/(1.228 \times 10^7)^2 = 2.64 m s2^{-2}. Centripetal balance gives orbital speed v=gr=2.64×1.228×107=5694v = \sqrt{gr} = \sqrt{2.64 \times 1.228 \times 10^7} = 5694 m s1^{-1}, and period T=2πr/v=2π×1.228×107/5694=1.35×104T = 2\pi r/v = 2\pi \times 1.228 \times 10^7/5694 = 1.35 \times 10^4 s = 3.773.77 hours, matching the observed orbital period.

Example 2. Snowy 2.0 pumped-hydro gravitational PE storage. Snowy 2.0 lifts water 700700 m from Talbingo (lower) to Tantangara (upper). Treating gg as uniform at 9.89.8 m s2^{-2} over this 700700 m altitude range, gravitational PE stored per kg is ΔU=mgh=9.8×700=6860\Delta U = mgh = 9.8 \times 700 = 6860 J, or 1.911.91 kWh per tonne. Tantangara's 254254 GL (2.54×10112.54 \times 10^{11} kg) thus stores 2.54×1011×6860=1.74×10152.54 \times 10^{11} \times 6860 = 1.74 \times 10^{15} J or 485485 GWh of potential energy. Released through turbines, this powers 22002200 MW for 220\approx 220 hours, providing the seasonal balance for renewable variability.

Try this

Q1. State Newton's law of universal gravitation and define each symbol. [2 marks]

  • Cue. F=Gm1m2/r2F = Gm_1 m_2/r^2. GG is the gravitational constant; m1,m2m_1, m_2 are interacting masses; rr is centre-to-centre distance.

Q2. Earth-orbit radius RE=6.378×106R_E = 6.378 \times 10^6 m, ME=5.97×1024M_E = 5.97 \times 10^{24} kg. Calculate (a) gravitational field strength at surface, and (b) at altitude 400400 km (ISS orbit). [4 marks]

  • Cue. (a) GM/R2=9.80GM/R^2 = 9.80 m s2^{-2}. (b) r=6.778×106r = 6.778 \times 10^6, g=8.68g = 8.68 m s2^{-2}.

Q3. Refer to Snowy 2.0 PE storage. (a) Calculate the energy stored per kg lifted 700700 m. (b) Determine the total stored energy in 254254 GL of water. (c) Explain why gg may be treated as uniform over this height range. [2+3+2 marks]

  • Cue. (a) 68606860 J. (b) 1.74×10151.74 \times 10^{15} J. (c) Altitude change of 700700 m is small compared with Earth radius, so gg varies by less than 0.02%0.02\%.

Exam-style practice questions

Practice questions written in the style of VCAA exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

2023 VCE3 marksCalculate the gravitational field strength at the surface of Mars. Mass of Mars = 6.4 x 10^23 kg, radius of Mars = 3.4 x 10^6 m. (G = 6.67 x 10^-11 N m^2 / kg^2.)
Show worked answer →

The gravitational field strength at the surface of a planet is:

g=GMr2=6.67×1011×6.4×1023(3.4×106)2g = \frac{G M}{r^2} = \frac{6.67 \times 10^{-11} \times 6.4 \times 10^{23}}{(3.4 \times 10^6)^2}

Numerator: 6.67×1011×6.4×1023=4.27×10136.67 \times 10^{-11} \times 6.4 \times 10^{23} = 4.27 \times 10^{13}.

Denominator: (3.4×106)2=1.156×1013(3.4 \times 10^6)^2 = 1.156 \times 10^{13}.

g=4.27×10131.156×1013=3.7g = \frac{4.27 \times 10^{13}}{1.156 \times 10^{13}} = 3.7 N/kg.

Markers reward correct substitution, the squared radius in the denominator, and the unit N/kg (equivalent to m/s squared).

2025 VCE4 marksA 1500 kg satellite is moved from the surface of the Earth to an altitude of 600 km. Use the area under a force-distance graph to estimate the change in gravitational potential energy. Mass of Earth = 6.0 x 10^24 kg, radius of Earth = 6.4 x 10^6 m.
Show worked answer →

The change in gravitational potential energy equals the area under the FF vs rr graph from r1r_1 to r2r_2 (it is the work done against gravity moving outward).

r1=6.4×106r_1 = 6.4 \times 10^6 m; r2=6.4×106+6.0×105=7.0×106r_2 = 6.4 \times 10^6 + 6.0 \times 10^5 = 7.0 \times 10^6 m.

F1=GMmr12=6.67×1011×6.0×1024×1500(6.4×106)2=1.47×104F_1 = \frac{G M m}{r_1^2} = \frac{6.67 \times 10^{-11} \times 6.0 \times 10^{24} \times 1500}{(6.4 \times 10^6)^2} = 1.47 \times 10^4 N.

F2=GMmr22=6.67×1011×6.0×1024×1500(7.0×106)2=1.22×104F_2 = \frac{G M m}{r_2^2} = \frac{6.67 \times 10^{-11} \times 6.0 \times 10^{24} \times 1500}{(7.0 \times 10^6)^2} = 1.22 \times 10^4 N.

Approximating the area under the curve from r1r_1 to r2r_2 as a trapezium:

ΔU12(F1+F2)(r2r1)=12(1.47+1.22)×104×6.0×105\Delta U \approx \frac{1}{2} (F_1 + F_2) (r_2 - r_1) = \frac{1}{2} (1.47 + 1.22) \times 10^4 \times 6.0 \times 10^5

ΔU12×2.69×104×6.0×105=8.1×109\Delta U \approx \frac{1}{2} \times 2.69 \times 10^4 \times 6.0 \times 10^5 = 8.1 \times 10^9 J.

Markers accept the trapezium estimate, reward correct F1F_1 and F2F_2, and accept exact answers using ΔU=GMm(1/r11/r2)\Delta U = G M m (1/r_1 - 1/r_2) for comparison.

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